Question

Find the angle BCD and angle ECD

Answer 1

$\huge\bold{ANSWER~:}$
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( I denote angle as L )

L DAC = L DBC ( therome 10.9 )

L DAC = 70°

L BAD = L BAC + L DAC

L BAD = 30° + 70°

L BAD = 100°

In cyclic quadrialteral ABCD ,

L BAD + L BCD = 180° ( therome 10.11 )

100° + L BCD = 180°

$\boxed{L~BCD~=~80°}$

In ΔABC,

If AB = BC than, L BAC = L BCA

L BCA = 30°

L BCD = L BAC + L ACD

80° = 30° + L ACD

L ACD = 50°

$\boxed{~L~ECD~=~50°}$

${\large\bold\color{Red} { Answer~is~ L~BCD~=~80° }}$

${\large\bold\color{Red} { Answer~is~ L~ECD~=~50° }}$

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